Question 553901
<pre>
This problem has no real solutions. Let me explain what is 
going on here.  

The equation 

x² - 4y² = 16

if you were to solve it for y and plot a bunch of points, like these:

(-8.5,-3.75), (-8.5,3.75), (-5,-1.5), (-5,1.5), (-4,0)
(4,0), (5,-1.5), (5,1.5), (8.5,-3.75), (8.5,3.75)

you would get this graph, called a hyperbola.  It is called a "conic section"
because there are several kinds of curves that can be gotten by slicing a
cone different ways.  The other tutor is right that you need not be
concered about that at this time.  Anyway here is the graph of x² - 4y² = 16.

{{{drawing(400,200,-10,10,-5,5,
graph(400,200,-10,10,-5,5,sqrt(x^2-16)/2),
graph(400,200,-10,10,-5,5,-sqrt(x^2-16)/2) )}}}

Now let's look at the other equation, 

y = 3x - 3

If you were to plot some points, like these:
 
(0,-3), (1,0), (2,3)

you will get a straight line like thisL

{{{drawing(400,200,-10,10,-5,5,
graph(400,200,-10,10,-5,5,15,3x-3) )}}}

But if you put them both on the same graph, like this:

{{{drawing(400,200,-10,10,-5,5,
graph(400,200,-10,10,-5,5,sqrt(x^2-16)/2,3x-3),
graph(400,200,-10,10,-5,5,-sqrt(x^2-16)/2) )}}}

you find that they don't intersect at all!  That means there
cannot be any REAL point where they intersect. 

Suppose you did as the other tutor said:

x² - 4y² = 16
y = 3x - 3

Substitute 3x - 3 for y in the first equation:

x² - 4(3x - 3)² = 16
x² - 4(3x - 3)(3x - 3) = 16
x² - 4(9x² - 18x + 9) = 16
x² - 36x² + 72x - 36 = 16
-35x² + 72x - 36 = 16
-35x² + 72x - 52 = 0
 35x² - 72x + 52 = 0

That does not factor and if you use the quadratic formula:

x = {{{(-b +- sqrt( b^2-4*a*c ))/(2*a) }}}  
 
x = {{{(-(-72) +- sqrt( (-72)^2-4*(35)*(52) ))/(2*(35)) }}} 

x = {{{(72 +- sqrt(5184-7280 ))/(70) }}} 

x = {{{(72 +- sqrt(-2096))/70 }}}

x = {{{(72 +- i*sqrt(2096))/70 }}}

x = {{{(72 +- i*sqrt(16*131))/70 }}}

x = {{{(72 +- 4i*sqrt(131))/70 }}}

x = {{{(4(18 +- i*sqrt(131)))/70 }}}

x = {{{(2(18 +- i*sqrt(131)))/35 }}}

Which is an imaginary answer, not a real answer

----------------------------------------------

Now if the second equation had been 10y + 3x = 0
instead and your problem had been:

x² - 4y² = 16
10y + 3x = 0

and we found and plotted these points, say

(0,0), (4, -1.2), (7,-2.1) 

and had gotten this line instead, there would have
been two real solutions at the points where these two
graphs intersect, as shown with the two small circles. 

{{{drawing(400,200,-10,10,-5,5,
graph(400,200,-10,10,-5,5,sqrt(x^2-16)/2,.3x),
graph(400,200,-10,10,-5,5,-sqrt(x^2-16)/2),
circle(-5,-3/2,.3), circle(5,3/2,.3)


 )}}}

Suppose you did as the other tutor said in this
problem:

x² - 4y² = 16
10y = 3x

Solve the second equation for y

y = {{{3x/10}}}

Substitute {{{3x/10}}} for y in the first equation:
 
x² - 4y² = 16

x² - 4{{{(3x/10)^2}}} = 16

x² - 4{{{(9x^2/100)}}} = 16

x² - {{{(9x^2/25)}}} = 16

Multiply through by 25 to clear of fractions:

25x² - 9x² = 400
      16x² = 400
        x² = 25
         x = ±5

Using x = +5 you would have gotten

y = {{{3x/10}}}
y = {{{3(5)/10}}
y = {{{15/10}}}
y = {{{3/2}}}

So one solution would be the point (5,{{{3/2}}}) or (5,1.5)

Using x = -5 you would have gotten

y = {{{3x/10}}}
y = {{{3(-5)/10}}
y = {{{-15/10}}}
y = {{{-3/2}}}

So the other solution would be (-5,{{{-3/2}}}) or (-5,-1.5)

Notice that the points where the graphs intersect are indeed
those very points (5,1.5) and (-5,-1.5)

{{{drawing(400,200,-10,10,-5,5,
graph(400,200,-10,10,-5,5,sqrt(x^2-16)/2,.3x),
graph(400,200,-10,10,-5,5,-sqrt(x^2-16)/2),
circle(-5,-3/2,.3), circle(5,3/2,.3),
locate(1.5,2,"(5,1.5)"), locate(-4.5,-1.5,"(-5,-1.5)")

 )}}}

Edwin</pre>